NEET · Physics · STD 11 - 4.2 friction
A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in \(m s ^{-2}\) is :
\(\left(g=10 m / s ^2\right)\)
- A 2.1
- B 1.8
- C 1.5
- D 1.2
Answer & Solution
Correct Answer
(D) 1.2
Step-by-step Solution
Detailed explanation
(D) 1.2
Let the acceleration of the trolley be a.
In the frame of the trolley, a pseudo force ma acts on the box in the direction opposite to the motion.
For the box to remain stationary relative to the trolley, the static friction force must balance this pseudo force.
The maximum static friction force available is given by \(f_{s, \max }=\mu_s N=\mu_s m g\).
Equating the pseudo force to the maximum static friction force gives the maximum acceleration:
\(m a_{\max }=\mu_s m g\)
\(a_{\max }=\mu_s g\)
Substituting the given values \(\mu_s=0.12\) and \(g=10 m / s ^2\) :
\(a_{\max }=0.12 \times 10=1.2 m / s ^2\)
Let the acceleration of the trolley be a.
In the frame of the trolley, a pseudo force ma acts on the box in the direction opposite to the motion.
For the box to remain stationary relative to the trolley, the static friction force must balance this pseudo force.
The maximum static friction force available is given by \(f_{s, \max }=\mu_s N=\mu_s m g\).
Equating the pseudo force to the maximum static friction force gives the maximum acceleration:
\(m a_{\max }=\mu_s m g\)
\(a_{\max }=\mu_s g\)
Substituting the given values \(\mu_s=0.12\) and \(g=10 m / s ^2\) :
\(a_{\max }=0.12 \times 10=1.2 m / s ^2\)
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